# show that $\left(\sum_{i=0}^\infty\frac{a^i}{i!}\right)\left(\sum_{j=0}^\infty\frac{b^j}{j!}\right) = \sum_{k=0}^\infty\frac{(a+b)^k}{k!}$

I need to show that if $\sum_{i=0}^\infty \frac{a^i}{i!}$ is absolutely convergent for all $a\in\mathbb{R}$, then $$\left(\sum_{i=0}^\infty\frac{a^i}{i!}\right)\left(\sum_{j=0}^\infty\frac{b^j}{j!}\right) = \sum_{k=0}^\infty\frac{(a+b)^k}{k!}.$$

We see that the Cauchy product of the two series on the left hand side is defined to be $$\sum_{k=0}^\infty \sum_{i+j=k}^\infty \frac{a^i}{i!}\cdot \frac{b^j}{j!}.$$

I'm having trouble showing that $$\sum_{k=0}^\infty\sum_{i+j=k}^\infty \frac{a^i}{i!}\cdot \frac{b^j}{j!} = \sum_{k=0}^\infty\frac{(a+b)^k}{k!}$$

would anyone be able to help? I know that once I get this then since the two series on LHS on the very top are convergent, then so is the Cauchy product. Thus the statement would be proved.

We have that \begin{align*} \left(\sum_{i=0}^\infty\frac{a^i}{i!}\right)\left(\sum_{j=0}^\infty\frac{b^j}{j!}\right)&=\sum_{k=0}^{\infty}\sum_{(i,j)\in S_k}\frac{a^i}{i!}\cdot \frac{b^j}{j!}\\ &=\sum_{k=0}^{\infty}\sum_{(i,j)\in S_k}\frac{1}{(i+j)!} \binom{i+j}{i}a^ib^j\\ &=\sum_{k=0}^{\infty}\frac{1}{k!} \sum_{i=0}^k\binom{k}{i}a^ib^{k-i}=\sum_{k=0}^{\infty}\frac{(a+b)^k}{k!} \end{align*} where $S_k:=\{(i,j): i\geq 0, j\geq 0, i+j=k\}$.
Hint $\sum_\limits{i+j=k}$ doesn't mean $\sum_\limits{i+j=k}^\infty$ as you wrote. What it really means is
$$\sum_\limits{i+j=k} u_{i, j} = \sum_\limits{i=0}^k u_{i, k-i}$$